Mathematical knowledge: finding combinatorial number II - finding combinatorial number

subject ：AcWing 886. Find the combination number II
Given n Group inquiry , Each group is given two integers a,b, Please output C^b_amod(10⁹+7) Value .

Input format
The first line contains integers n.

Next n That's ok , Each line contains a set of a and b.

Output format
common n That's ok , Each line outputs a solution to the query .

Data range
1≤n≤10000,
1≤b≤a≤10⁵

sample input ：

3
3 1
5 3
2 2

sample output ：

3
10
1

Topic analysis ：
This problem is a preprocessed factorial
Because combinations can be replaced by factorial forms

But the division should be replaced by the inverse , So we need to use the knowledge of fast power to find the inverse element .
Last fact[i] Express i The factorial ,infact[i] Express i Inverse element
C_a^b=fact[a] * infact[a-b] * infact[b]

#include <iostream>

using namespace std;
typedef long long LL;
const int N = 100010,mod=1e9+7;

int fact[N],infact[N];

int qmi(int a, int k, int p)
{
    
    int res = 1;
    while(k)
    {
    
        if(k&1)res=(LL)res*a%p;
        a=(LL)a*a%p;
        k>>=1;
    }
    return res;
}
int main()
{
    
    fact[0]=infact[0]=1;
    for(int i=1;i<N;i++) //  Preprocessing 
    {
    
        fact[i]=(LL)fact[i-1]*i%mod; //  Factorial 
        infact[i]=(LL)infact[i-1]*qmi(i,mod-2,mod)%mod; //  Inverse of factorial 
    }
    
    int n;
    cin>>n;
    while(n--)
    {
    
        int a,b;
        cin>>a>>b;
        cout<<(LL)fact[a]*infact[b]%mod*infact[a-b]%mod<<endl;
    }
    return 0;
}